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2 years ago

Given the points X(-2, 5) and Y(2, -3), find the coordinates of the point P on

Mathematics

1 answer:

Alinara [238K]2 years ago

5 0

The coordinate of the point P will be (6/5, -7/5). Then the correct option is C.

The complete question is attached below.

<h3>What is the section of the line?</h3>

Let A (x₁, y₁) and B (x₂, y₂) be a line segment. Then the point P (x, y) divide the line segment in the ratio of m:n. Then we have

x = (mx₂ + nx₁) / (m + n)

y = (my₂ + ny₁) / (m + n)

Given the points X(-2, 5) and Y(2, -3).

Then the coordinates of the point P on directed line segment XY that partitions segment XY such that the ratio of XP to PY is 4:1.

(x₁, y₁) = (-2, 5)

(x₂, y₂) = (2, -3)

m = 4

n = 1

Then the point will be

x = [4 × 2 + 1 × (-2)] / (4 + 1)

x = 6 / 5

y = [4 × (-3) + 1 × 5] / (4 + 1)

y = -7 / 5

The coordinate of the point P will be (6/5, -7/5).

Then the correct option is C.

More about the section of the line link is given below.

brainly.com/question/6582647

#SPJ1

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We want to verify the equation:

$\displaystyle \frac{1}{\sec\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

We can convert sec(α) to 1 / cos(α):

$\displaystyle \frac{1}{1/\cos\alpha+1}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

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$\displaystyle \frac{\cos\alpha}{1+\cos\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Create a common denominator. We can multiply the first fraction by (1 - cos(α)):

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{(1+\cos\alpha)(1-\cos\alpha)}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Simplify:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{1-\cos^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

From the Pythagorean Identity, we know that cos²(α) + sin²(α) = 1 or equivalently, 1 - cos²(α) = sin²(α). Substitute:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)}{\sin^2\alpha}-\frac{\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha }{\sin^2\alpha }-\frac{1}{\sec\alpha -1}$

Subtract:

$\displaystyle \frac{\cos\alpha(1-\cos\alpha)-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Distribute:

$\displaystyle \frac{\cos\alpha-\cos^2\alpha-\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Rewrite:

$\displaystyle \frac{(\cos\alpha)-(\cos^2\alpha+\cos\alpha)}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Split:

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos^2\alpha+\cos\alpha}{\sin^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor the second fraction, and substitute sin²(α) for 1 - cos²(α):

$\displaystyle \frac{\cos\alpha}{\sin^2\alpha}-\frac{\cos\alpha(\cos\alpha+1)}{1-\cos^2\alpha}=\frac{\cos\alpha}{\sin^2\alpha}-\frac{1}{\sec\alpha-1}$

Factor:

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Cancel:

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Divide the second fraction by cos(α):

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